leetcode Range Sum Query - Mutable

原题链接：https://leetcode.com/problems/range-sum-query-mutable/

Description

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

The update(i, val) function modifies nums by updating the element at index i to val.

Example:

Given nums = [1, 3, 5]

sumRange(0, 2) -> 9

update(1, 2)

sumRange(0, 2) -> 8

Note:

The array is only modifiable by the update function.

You may assume the number of calls to update and sumRange function is distributed evenly.

线段树单点更新。。

class NumArray {
public:
NumArray() = default;
NumArray(vector<int> &nums) {
n = (int)nums.size();
if (!n) return;
arr = new int[n << 2];
built(1, 1, n, nums);
}
~NumArray() { delete []arr; }
void update(int i, int val) {
update(1, 1, n, i + 1, val);
}
int sumRange(int i, int j) {
return sumRange(1, 1, n, i + 1, j + 1);
}
private:
int n, *arr;
void built(int root, int l, int r, vector<int> &nums) {
if (l == r) {
arr[root] = nums[l - 1];
return;
}
int mid = (l + r) >> 1;
built(root << 1, l, mid, nums);
built(root << 1 | 1, mid + 1, r, nums);
arr[root] = arr[root << 1] + arr[root << 1 | 1];
}
void update(int root, int l, int r, int pos, int val) {
if (pos > r || pos < l) return;
if (pos <= l && pos >= r) {
arr[root] = val;
return;
}
int mid = (l + r) >> 1;
update(root << 1, l, mid, pos, val);
update(root << 1 | 1, mid + 1, r, pos, val);
arr[root] = arr[root << 1] + arr[root << 1 | 1];
}
int sumRange(int root, int l, int r, int x, int y) {
if (x > r || y < l) return 0;
if (x <= l && y >= r) return arr[root];
int mid = (l + r) >> 1, ret = 0;
ret += sumRange(root << 1, l, mid, x, y);
ret += sumRange(root << 1 | 1, mid + 1, r, x, y);
return ret;
}
};